The total number of matrices $A=\begin{bmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{bmatrix}$,…
The total number of matrices $A=\begin{bmatrix} 0 & 2y & 1 \\ 2x & y & -1 \\ 2x & -y & 1 \end{bmatrix}$, where $x, y \in \mathbb{R}$, and $x \neq y$ for which $A^TA = 3I_3$ is:
Solution
Given $A^TA = 3I_3$
$\begin{aligned}
\begin{vmatrix}
0 & 2x & 2x \\
2y & y & -y \\
1 & -1 & 1 \\
\end{vmatrix}
\begin{vmatrix}
0 & 2y & 1 \\
2x & y & -1 \\
2x & -y & 1 \\
\end{vmatrix}
=
\begin{vmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{vmatrix}
\end{aligned}$
$\begin{aligned}
\begin{vmatrix}
8x^2 & 0 & 0 \\
0 & 6y^2 & 0 \\
0 & 0 & 3 \\
\end{vmatrix}
=
\begin{vmatrix}
3 & 0 & 0 \\
0 & 3 & 0 \\
0 & 0 & 3 \\
\end{vmatrix}
\end{aligned}$
On comparing the two matrices, we get
$8x^2 = 3$ and $6y^2 = 3$
$\Rightarrow x^2 = \frac{3}{8}$ and $y^2 = \frac{1}{2}$
$\Rightarrow x = \pm \sqrt{\frac{3}{8}} and y = \pm \sqrt{\frac{1}{2}}$
For each of $x$ and $y$ we have two possible values, hence there are total $2 \times 2 = 4$ combinations of values are possible.
Hence, total $4$ matrices are possible.